Question: You have found the following ages (in years) of all 5 snakes at your local zoo: $ 10,\enspace 20,\enspace 21,\enspace 8,\enspace 7$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{10 + 20 + 21 + 8 + 7}{{5}} = {13.2\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $-3.2$ years $10.24$ years $^2$ $20$ years $6.8$ years $46.24$ years $^2$ $21$ years $7.8$ years $60.84$ years $^2$ $8$ years $-5.2$ years $27.04$ years $^2$ $7$ years $-6.2$ years $38.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{10.24} + {46.24} + {60.84} + {27.04} + {38.44}} {{5}} $ $ {\sigma^2} = \dfrac{{182.8}}{{5}} = {36.56\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{36.56\text{ years}^2}} = {6\text{ years}} $ The average snake at the zoo is 13.2 years old. There is a standard deviation of 6 years.